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Basic Thermodynamics

[ "article:topic", "Author tag:Tuckerman", "showtoc:no" ]
  • Page ID
    5168
  • In the microcanonical ensemble, the entropy \(S\) is a natural function of \(N\),\(V\) and \(E\), i.e., \(S=S(N,V,E)\). This can be inverted to give the energy as a function of \(N\), \(V\), and \(S\), i.e., \(E=E(N,V,S)\). Consider using Legendre transformation to change from \(S\) to \(T\) using the fact that

    \[T= \left(\frac {\partial E}{\partial S}\right)_{N,V}\]

    The Legendre transform \(\tilde{E}\) of \(E(N,V,S)\) is

    \[ \tilde {E} (N, V, T ) = E (N,V,S(T)) - S \frac {\partial E}{\partial S}\]

    \[ = E(N,V,S(T)) - TS \]

    The quantity \(\tilde{E}(N,V,T)\) is called the Hemlholtz free energy and is given the symbol \(A(N,V,T)\) and is the fundamental energy in the canonical ensemble.
    The differential of \(A\) is

    \[dA = \left( \partial A \over \partial T \right)_{N,V}dT + \left( \partial A \over \partial V \right)_{N,T} dV +\left( \partial A \over \partial N \right)_{T,V} dN\]

    However, from \(A = E - TS \), we have

    \[ dA = dE - TdS - SdT \]

    From the first law, \(dE\) is given by

    \[ dE = TdS - PdV + \mu dN \]

    Thus,

    \[ dA = - PdV - S dT + \mu dN \]

    Comparing the two expressions, we see that the thermodynamic relations are

    \[ S =  -\left(\frac {\partial A}{\partial T}\right)_{N,V}\]

    \[ P =  -\left(\frac {\partial A}{\partial V}\right)_{N,T}\]

    \[ \mu =  -\left(\frac {\partial A}{\partial N}\right)_{V,T}\]